Ncert Solutions Physics Class 12th

6.10 Speed of the jet plane, v = 1800 km/h = 500 m/s

Wing span, l = 25 m

Earth's magnetic field, B = 5 $*{10}^{-4}$ T

Dip angle, $\beta$ = 30 $°$

The vertical component of the earth's magnetic field, $B_V$ = B $\sin ?\beta$ = 5 $*{10}^{-4}\sin ?30°$= 2.5 $*{10}^{-4}$ T

Voltage difference between the ends of the wing can be calculated as

e = $B_V*l*v$ = 2.5 $*{10}^{-4}*25*500$ = 3.125 V

6.9 Mutual inductance of a pair of coils, $\mu$ = 1.5 H

Initial current, $I_1$ = 0 A

Final current, $I_2$ = 20 A

Change in current, dI = $I_2$ - $I_1$ = 20 A

Time taken for change, dt = 0.5 s

From the relation of induced emf e = $\frac{d\varnothing }{dt}$ , where $d\varnothing$ is the change in the flux linkage with the coil ……………(1)

The relation between emf and mutual inductance is e = $\mu \frac{dI}{dt}$ …………. (2)

Combining equations (1) and (2), we get

$\frac{d\varnothing }{dt}=\mu \frac{dI}{dt}$ or $d\varnothing =\mu dI$ = 1.5 $*20$ = 30 Wb

Hence, the change in the flux linkage is 30 Wb

6.8 Given:

Initial current,  $I_1$ = 5.0 A

Final current,  $I_2$ = 0 A

Change in current, di = $I_1-I_2$ = 5 A

Time taken for the change, dt = 0.1 s

Average emf, e = 200 V

For self-inductance (L) of the coil, L is given by

L = $\frac{e}{\frac{di}{dt}}$ = $\frac{200}{\frac{5}{0.1}}$ = 4 H

Hence, the self-inductance of the coil is 4 H.

As per the NCERT Textbooks

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These ICs are used in majority of our daily use electronics and all advance electronics. 

2.35 According to Gauss's law, the electric field between a sphere and a shell is determined by the charge $q_1$ on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge $q_2$ . For positive charge $q_1$ , potential difference V s always positive.

2.34 Equidistant planes parallel to the x-y plane are the equipotential surfaces.

Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases

Concentric spheres centered at the origin are equipotential surfaces.

A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.

2.32  Length of the co-axial cylinder, l = 15 cm = 0.15 m

Radius of the outer cylinder, $r_1$ = 1.5 cm = 0.015 m

Radius of the inner cylinder, $r_2$ = 1.4 cm = 0.014 m

Charge on the inner cylinder, q = 3.5 $\mu C=3.5*{10}^{-6}$ C

Capacitance of a co-axial cylinder of radii $r_1$ and $r_2$ is given by the relation

C = $\frac{2\pi {\epsilon }_{0}l}{{\log }_e?\frac{r_1}{r_2}}$ , where ${\epsilon }_0$ = permittivity of free space = 8.854 $*{10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

C = $\frac{2\pi *8.854\mathrm{}*{10}^{-12}*0.15}{{\log }_e?\frac{0.015}{0.014}}$ = 1.21 $*{10}^{-10}$ F

Potential difference of the inner cylinder is given by

V = $\frac{q}{C}$ = $\frac{3.5*{10}^{-6}}{1.21\mathrm{}*{10}^{-10}}$ = 2.893 $*{10}^4$ V